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| + | ===== Deformed natural ===== | ||
| + | ==== Function ==== | ||
| + | | @#55CCEE: context | @#55CCEE: $p\in Q$ | | ||
| + | | @#55CCEE: context | @#55CCEE: $u:{\mathbb N}\to{}Q\to{\mathbb A}$ | | ||
| + | | @#FF9944: definition | @#FF9944: $[n]_u(q) := \sum_{k=1}^n \dfrac{u_k(q)}{u_k(p)}$ | | ||
| + | And clearly the denominator must be nonzero. | ||
| + | |||
| + | === Discussion === | ||
| + | E.g., for another sequence $a_n$ consider $u(n,q):=q^{a_n}$. | ||
| + | |||
| + | In particular, consider $a_n:=n*x+d$ for some $d$. | ||
| + | |||
| + | <code> | ||
| + | a[k_] = k x + d; | ||
| + | Sum[q^a[k], {k, 1, n}] | ||
| + | Limit[%, q -> 1] | ||
| + | </code> | ||
| + | |||
| + | In particular, consider $x:=1, d:=0$ for [[quantum_integer]]s. | ||
| + | |||
| + | === Theorems === | ||
| + | $\lim_{q\to p}[n]_u(q) = \sum_{k=1}^n 1 = n$ | ||
| + | |||
| + | Also, for any sequence $(a_k)$, | ||
| + | |||
| + | $\sum_{k=1}^n a_k = \lim_{q\to 1}\dfrac{\partial}{\partial{}q} \sum_{k=1}^n q^{a_k}$ | ||
| + | |||
| + | === Reference === | ||
| + | Wikipedia: [[http://en.wikipedia.org/wiki/Q-analog|q-analog]] | ||
| + | |||
| + | ----- | ||
| + | === Requirements === | ||
| + | [[Metric space]] | ||
| + | === Related === | ||
| + | [[quantum_integer]] | ||
